| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2014 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sign Change & Interval Methods |
| Type | Interval Bisection with Other Methods |
| Difficulty | Standard +0.3 This is a straightforward numerical methods question requiring routine application of sign change verification, linear interpolation, and interval bisection. All three parts follow standard algorithms with no conceptual challenges—students simply execute the procedures they've learned. The calculations are mechanical and the question structure is typical for FP1. |
| Spec | 1.09a Sign change methods: locate roots1.09b Sign change methods: understand failure cases |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f(2) = -1.9609...\), \(f(3) = 3.8805...\) | M1 | Attempts to evaluate both \(f(2)\) and \(f(3)\), evaluates at least one correctly to awrt (or trunc.) 2 sf |
| Sign change (and \(f(x)\) is continuous) therefore a root \(\alpha\) is between \(x=2\) and \(x=3\) | A1 | Both values correct to awrt 2 sf, sign change stated, and conclusion |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{\alpha - 2}{\text{"1.9609..."}} = \frac{3 - \alpha}{\text{"3.8805..."}}\) | M1 | Correct linear interpolation method, must be a correct statement using their \(f(2)\) and \(f(3)\). If any "negative lengths" are used, score M0 |
| \(\alpha_2 = \frac{3 \times 1.96.. + 2 \times 3.88..}{1.96...+3.88...}\) | A1ft | Follow through their values if seen explicitly |
| \(\alpha_2 = 2.336\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f(0) = +(1)\) or \(f(-1) = -(4.248)\) | B1 | Award for correct sign, can be in a table |
| \(f(-0.5) = -0.879...\) | M1 | Attempt \(f(-0.5)\) |
| \(f(-0.25) = 0.382...\) | M1 | Attempt \(f(-0.25)\) |
| \(\therefore -0.5 < \beta < -0.25\) | A1 | oe with no numerical errors seen |
## Question 2:
$f(x) = 3\cos 2x + x - 2$
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(2) = -1.9609...$, $f(3) = 3.8805...$ | M1 | Attempts to evaluate both $f(2)$ and $f(3)$, evaluates at least one correctly to awrt (or trunc.) 2 sf |
| **Sign change** (and $f(x)$ is continuous) therefore a **root** $\alpha$ is between $x=2$ and $x=3$ | A1 | Both values correct to awrt 2 sf, sign change stated, and conclusion |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{\alpha - 2}{\text{"1.9609..."}} = \frac{3 - \alpha}{\text{"3.8805..."}}$ | M1 | Correct linear interpolation method, must be a correct statement using their $f(2)$ and $f(3)$. If any "negative lengths" are used, score M0 |
| $\alpha_2 = \frac{3 \times 1.96.. + 2 \times 3.88..}{1.96...+3.88...}$ | A1ft | Follow through their values if seen explicitly |
| $\alpha_2 = 2.336$ | A1 | cao |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(0) = +(1)$ **or** $f(-1) = -(4.248)$ | B1 | Award for correct sign, can be in a table |
| $f(-0.5) = -0.879...$ | M1 | Attempt $f(-0.5)$ |
| $f(-0.25) = 0.382...$ | M1 | Attempt $f(-0.25)$ |
| $\therefore -0.5 < \beta < -0.25$ | A1 | oe with no numerical errors seen |
---2.
$$\mathrm { f } ( x ) = 3 \cos 2 x + x - 2 , \quad - \pi \leqslant x < \pi$$
\begin{enumerate}[label=(\alph*)]
\item Show that the equation $\mathrm { f } ( x ) = 0$ has a root $\alpha$ in the interval [2,3].\\[0pt]
\item Use linear interpolation once on the interval [2,3] to find an approximation to $\alpha$.
Give your answer to 3 decimal places.
\item The equation $\mathrm { f } ( x ) = 0$ has another root $\beta$ in the interval $[ - 1,0 ]$. Starting with this interval, use interval bisection to find an interval of width 0.25 which contains $\beta$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP1 2014 Q2 [9]}}